Question

Find the EMF of the unknown voltage source if a potentiometer has a one metre length. It has an unknown EMF voltage source balanced at $60cm$ length of potentiometer wire, while a $3V$ battery is balanced at $45cm$ length.
A) $3V$
B) $2.25V$
C) $4V$
D) $4.5V$

Answer 1

Hint: To find the unknown EMF of unknown voltage source we have to derive the expression which has EMF and balancing length. Then, divide unknown EMF with the known EMF by using the derived formula.

Complete step by step solution:
The potentiometer can be defined as the three – terminal resistor that has a voltage divider which is adjustable. It has sliding or rotating contact. When there are only two terminals wiper and one end then, the potentiometer can be used as rheostat or variable resistor.
Now, the relation between EMF and length of potentiometer wire can be derived as –
Using Ohm’s law, we get –
$V = IR$
Where, $V$ is the voltage
$I$ is the current
$R$ is the resistance
We know that,
$
  R = \rho \dfrac{L}{A}V \\
  R = I\rho \dfrac{L}{A} \\
 $
where, $\rho $ is the resistivity
$A$ is the cross – sectional area
Resistivity and cross – sectional area both are constant and in case of rheostat current is also constant.
$
  \therefore \dfrac{{L\rho }}{A} = KV = KLE \\
   \Rightarrow \dfrac{{L\rho l}}{A} = Kl \\
 $
where, $l$ is the balancing lengths
$E$ is the cell of unknown EMF
$K$ is the constant value
Due to flow of current equal to zero the potential difference also becomes equal to zero which gives null detection in the galvanometer. So, the unknown EMF can be calculated by knowing $K$ and $l$.
$
  E = \dfrac{{L\rho l}}{A} \\
  E = Kl \\
 $
Now, we got the relation between the EMF and balancing length of the potentiometer.
Now, let the unknown EMF be ${E_1}$ and known EMF be ${E_2}$ and let the balancing lengths be ${l_1}$ and ${l_2}$ respectively –
$
  \therefore {E_1} = K{l_1} \cdots \left( 1 \right) \\
  {E_2} = K{l_2} \cdots \left( 2 \right) \\
 $
Dividing equation $\left( 1 \right)$ from equation $\left( 2 \right)$, we get –
$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{l_1}}}{{{l_2}}} \cdots \left( 3 \right)$
According to the question, it is given that –
$
  {E_2} = 3V \\
  {l_1} = 60cm \\
  {l_2} = 35cm \\
 $
Putting these values in equation $\left( 3 \right)$, we get –
$\dfrac{{{E_1}}}{3} = \dfrac{{60}}{{35}}$
Further solving, we get –
${E_1} = 4V$
Therefore, we got the value of unknown EMF as $4V$

Hence, the correct option is (C).

Note: The potentiometer has a known EMF $V$. Its voltage is also known as driver cell voltage. This consists of long resistive wire of $L$. The two ends of $L$ are connected with the battery terminals. The cell whose EMF is to be measured is connected with the primary circuit and the other end is connected to the galvanometer $G$. It depends upon the principle that potential difference across any portion of wire which is directly proportional to the length of wire that has uniform cross – section area and current is constant.